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Let a, b, c, d be positive real numbers such that abcd = 1 and a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}. Prove that
a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}
Proposed by Pavel Novotný, Slovakia

Slični zadaci

Prove that for any four positive real numbers a, b, c, d the inequality
\frac {(a - b)(a - c)}{a + b + c} + \frac {(b - c)(b - d)}{b + c + d} + \frac {(c - d)(c - a)}{c + d + a} + \frac {(d - a)(d - b)}{d + a + b} \geqslant 0
holds. Determine all cases of equality.

Author: Darij Grinberg (Problem Proposal), Christian Reiher (Solution), Germany
Let S\subseteq\mathbb{R} be a set of real numbers. We say that a pair (f, g) of functions from S into S is a Spanish Couple on S, if they satisfy the following conditions:
(i) Both functions are strictly increasing, i.e. f(x) < f(y) and g(x) < g(y) for all x, y\in S with x < y;

(ii) The inequality f\left(g\left(g\left(x\right)\right)\right) < g\left(f\left(x\right)\right) holds for all x\in S.

Decide whether there exists a Spanish Couple on the set S = \mathbb{N} of positive integers; on the set S = \{a - \frac {1}{b}: a, b\in\mathbb{N}\}

Proposed by Hans Zantema, Netherlands
Let a_1, a_2, \ldots, a_{100} be nonnegative real numbers such that a^2_1 + a^2_2 + \ldots + a^2_{100} = 1. Prove that
a^2_1 \cdot a_2 + a^2_2 \cdot a_3 + \ldots + a^2_{100} \cdot a_1 < \frac {12}{25}.
Author: Marcin Kuzma, Poland
Prove the inequality:

\displaystyle \sum_{i < j}{\frac {a_{i}a_{j}}{a_{i} + a_{j}}}\leq \frac {n}{2(a_{1} + a_{2} + ... + a_{n})}\cdot \sum_{i < j}{a_{i}a_{j}}

for positive reals a_{1}, a_{2}, ..., a_{n}.
The sequence c_{0}, c_{1}, . . . , c_{n}, . . . is defined by c_{0}= 1, c_{1}= 0, and c_{n+2}= c_{n+1}+c_{n} for n \geq 0. Consider the set S of ordered pairs (x, y) for which there is a finite set J of positive integers such that x=\sum_{j \in J}{c_{j}}, y=\sum_{j \in J}{c_{j-1}}. Prove that there exist real numbers \alpha, \beta, and M with the following property: An ordered pair of nonnegative integers (x, y) satisfies the inequality m < \alpha x+\beta y < M if and only if (x, y) \in S.


Remark: A sum over the elements of the empty set is assumed to be 0.
Let n be an integer,n \geq 3. Let x_1, x_2, \ldots, x_n be real numbers such that x_i < x_{i+1} for 1 \leq i \leq n - 1. Prove that

\frac{n(n-1)}{2}\sum_{i < j}x_{i}x_{j}>\left(\sum^{n-1}_{i=1}(n-i)\cdot x_{i}\right)\cdot\left(\sum^{n}_{j=2}(j-1)\cdot x_{j}\right)